Circles: CBSE Class 10 Previous Year Questions

231 different questions from Circles (NCERT Chapter 10) asked in CBSE Class 10 Maths board exams 2022–2026. 8 of them came up in more than one year. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions100 questions 2 marks questions40 questions 3 marks questions52 questions 4 marks questions29 questions 5 marks questions10 questions

Most asked Circles questions

Q28 (OR) (OR)3 marksShort AnswerCircles CBSE 2023 · Basic 430/2/1

Prove that the parallelogram circumscribing a circle is a rhombus.

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Answer: Proved.

Let parallelogram ABCD touch the circle at P, Q, R, S on AB, BC, CD, DA.
Equal tangents from external points: AP = AS, BP = BQ, CR = CQ, DR = DS
Adding: AB + CD = AD + BC
In a parallelogram AB = CD and AD = BC, so 2AB = 2AD, i.e. AB = AD.
Thus AB = BC = CD = DA, so ABCD is a rhombus.

Also asked in: 2025 Standard 30/2/1, 2025 Standard 30/2/2, 2025 Standard 30/2/3, 2025 Basic 430/4/2, 2024 Standard 30/4/1, 2024 Standard 30/4/2, 2024 Standard 30/4/3, 2024 Standard 30/5/1, 2024 Standard 30/5/2, 2024 Standard 30/5/3, 2024 Basic 430/5/1, 2024 Basic 430/5/2, 2024 Basic 430/5/3, 2023 Basic 430/2/2, 2023 Basic 430/2/3

Q283 marksShort AnswerCircles CBSE 2023 · Standard 30/4/1

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

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Answer: Proved.

Let PA and PB be tangents from external point P to a circle with centre O, touching it at A and B.
A radius is perpendicular to the tangent at the point of contact, so $∠ O A P = ∠ O B P = 9 0^{\circ}$ .
In quadrilateral OAPB, the angles add up to $36 0^{\circ}$ :
$∠ A P B + ∠ A O B + 9 0^{\circ} + 9 0^{\circ} = 36 0^{\circ}$
$∠ A P B + ∠ A O B = 18 0^{\circ}$
Hence the angle between the tangents is supplementary to $∠ A O B$ .

Also asked in: 2025 Standard 30/2/1, 2025 Standard 30/2/2, 2025 Standard 30/2/3, 2025 Basic 430/4/1, 2025 Basic 430/4/2, 2025 Basic 430/4/3, 2024 Basic 430/2/1, 2024 Basic 430/2/2, 2024 Basic 430/2/3, 2023 Standard 30/4/2

Q242 marksVery Short AnswerCircles CBSE 2023 · Standard 30/5/3

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

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Answer: Proved.

Let AB be a diameter of a circle with centre O, and let PQ and RS be the tangents at A and B.
The tangent at any point is perpendicular to the radius through that point, so OA $⊥$ PQ and OB $⊥$ RS.
So $∠ O A P = 9 0^{\circ}$ and $∠ O B S = 9 0^{\circ}$ (P and S on opposite sides of AB).
These are alternate angles made by the transversal AB with PQ and RS, and they are equal.
Hence PQ $∥$ RS.

Also asked in: 2025 Standard 30/3/1, 2025 Standard 30/3/2, 2025 Standard 30/3/3, 2025 Basic 430/4/1, 2025 Basic 430/4/2, 2025 Basic 430/4/3, 2024 Standard 30/5/1, 2024 Standard 30/5/2, 2024 Standard 30/5/3

Q12 (OR) (OR)4 marksLong AnswerCircles CBSE 2022 · Basic 430/2/2

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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Answer: Proved.

Let quadrilateral ABCD circumscribe a circle with centre O, touching AB, BC, CD, DA at P, Q, R, S.
Join OA, OB, OC, OD, OP, OQ, OR, OS.
In $△ O A P$ and $△ O A S$ : OP = OS (radii), OA common, $∠ O P A = ∠ O S A = 9 0^{\circ}$ , so they are congruent (RHS) and $∠ A O P = ∠ A O S$ .
Similarly $∠ B O P = ∠ B O Q$ , $∠ C O Q = ∠ C O R$ , $∠ D O R = ∠ D O S$ .
The eight angles at O add up to $36 0^{\circ}$ , so $2 (∠ A O P + ∠ B O P + ∠ C O R + ∠ D O R) = 36 0^{\circ}$ .
Thus $∠ A O B + ∠ C O D = 18 0^{\circ}$ , and similarly $∠ B O C + ∠ A O D = 18 0^{\circ}$ .

Also asked in: 2026 Basic 430/4/3, 2025 Standard 30/1/1, 2025 Standard 30/1/2, 2025 Standard 30/1/3

Q124 marksLong AnswerCircles CBSE 2022 · Standard 30/2/3

Prove that a parallelogram circumscribing a circle is a rhombus.

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Answer: Proved.

Let parallelogram ABCD touch the circle at P, Q, R, S on AB, BC, CD, DA respectively.
Tangents from an external point are equal: AP = AS, BP = BQ, CR = CQ, DR = DS.
Adding: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ), i.e. AB + CD = AD + BC.
In a parallelogram AB = CD and AD = BC, so 2AB = 2AD, i.e. AB = AD.
Thus all four sides are equal: AB = BC = CD = DA.
Hence ABCD is a rhombus.

Also asked in: 2026 Basic 430/4/1, 2023 Standard 30/6/3, 2022 Basic 430/2/1, 2022 Basic 430/2/3

Q293 marksShort AnswerCircles CBSE 2023 · Standard 30/1/1

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $∠ P T Q = 2∠ O P Q$ .

Diagram for CBSE 2023 Class 10 Maths question 29

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Answer: Proved.

Let $∠ P T Q = θ$ .
TP = TQ (tangents from an external point), so $∠ T P Q = ∠ T QP = \frac{18 0 ^{\circ} - θ}{2} = 9 0^{\circ} - \frac{θ}{2}$ .
OP $⊥$ TP, so $∠ O P T = 9 0^{\circ}$ .
$∠ O P Q = 9 0^{\circ} - ∠ T P Q = 9 0^{\circ} - (9 0^{\circ} - \frac{θ}{2}) = \frac{θ}{2}$ .
Hence $∠ P T Q = 2∠ O P Q$ .

Also asked in: 2026 Standard 30/2/1, 2026 Standard 30/2/3, 2026 Basic 430/4/2, 2023 Standard 30/1/2, 2023 Standard 30/1/3, 2023 Standard 30/6/1, 2023 Standard 30/6/2

Q124 marksLong AnswerCircles CBSE 2022 · Basic 430/3/2

Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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Answer: Proved.

Let ABCD circumscribe a circle with centre O, touching AB, BC, CD, DA at P, Q, R, S.
Join OA, OB, OC, OD, OP, OQ, OR, OS.
$△ O A P ≅ △ O A S$ (OP = OS radii, AP = AS tangents, OA common; SSS), so $∠ A O P = ∠ A O S$ .
Similarly $∠ B O P = ∠ B O Q$ , $∠ C O Q = ∠ C O R$ , $∠ D O R = ∠ D O S$ .
Sum of angles at O: $2 (∠ A O P + ∠ B O P + ∠ C O R + ∠ D O R) = 36 0^{\circ}$
So $(∠ A O P + ∠ B O P) + (∠ C O R + ∠ D O R) = 18 0^{\circ}$ , i.e. $∠ A O B + ∠ C O D = 18 0^{\circ}$ .
Similarly $∠ B O C + ∠ A O D = 18 0^{\circ}$ . Hence opposite sides subtend supplementary angles at the centre.

Also asked in: 2025 Basic 430/4/1, 2025 Basic 430/4/3, 2022 Basic 430/4/1, 2022 Basic 430/4/2, 2022 Basic 430/4/3

Q313 marksShort AnswerCircles CBSE 2023 · Basic 430/4/1

Prove that the lengths of tangents drawn from an external point to a circle are equal.

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Answer: Proved.

Let PQ and PR be tangents from an external point P to a circle with centre O, touching it at Q and R. Join OQ, OR and OP.
$∠ O QP = ∠ O R P = 9 0^{\circ}$ (radius is perpendicular to tangent).
In right triangles OQP and ORP: OQ = OR (radii) and OP = OP (common).
So $△ O QP ≅ △ O R P$ (RHS).
Hence PQ = PR (CPCT).